Question

find the Fourier series​

Answer 1

The Fourier series of a function f(x) in the interval [N, N+2L] or (N, N+2L) is given by,

$\displaystyle\small\longrightarrow f(x)=\dfrac{a_0}{2}+\sum_{n=1}^\infty\left(a_n\cos\left(\dfrac{n\pi}{L}\,x\right)+b_n\sin\left(\dfrac{n\pi}{L}\,x\right)\right)$

where,

• $\displaystyle\smalla_0=\dfrac{1}{L}\int\limits_N^{N+2L}f(x)\ dx$

• $\displaystyle\smalla_n=\dfrac{1}{L}\int\limits_N^{N+2L}f(x)\cos\left(\dfrac{n\pi}{L}\,x\right)\ dx$

• $\displaystyle\smallb_n=\dfrac{1}{L}\int\limits_N^{N+2L}f(x)\sin\left(\dfrac{n\pi}{L}\,x\right)\ dx$

Here,

• $\displaystyle\smallf(x)=\dfrac{\pi-x}{2}$

and the interval is (0, 2), i.e., N = 0 and L = 1.

Finding $\displaystyle\smalla_0,$

$\displaystyle\small\longrightarrow a_0=\dfrac{1}{L}\int\limits_N^{N+2L}f(x)\ dx$

$\displaystyle\small\longrightarrow a_0=\int\limits_0^2\left(\dfrac{\pi-x}{2}\right)\ dx$

$\displaystyle\small\longrightarrow a_0=\dfrac{\pi}{2}\big[x\big]_0^2-\dfrac{1}{4}\left[x^2\right]_0^2$

$\displaystyle\small\longrightarrow a_0=\pi-1$

FInding $\displaystyle\smalla_n,$

$\displaystyle\small\longrightarrow a_n=\dfrac{1}{L}\int\limits_N^{N+2L}f(x)\cos\left(\dfrac{n\pi}{L}\,x\right)\ dx$

$\displaystyle\small\longrightarrow a_n=\int\limits_0^2\left(\dfrac{\pi-x}{2}\right)\cos(n\pi x)\ dx$

$\displaystyle\small\longrightarrow a_n=\dfrac{\pi}{2}\int\limits_0^2\cos(n\pi x)\ dx-\dfrac{1}{2}\int\limits_0^2x\cos(n\pi x)\ dx$

$\displaystyle\small\longrightarrow a_n=\dfrac{\pi}{2n\pi}\big[\sin(n\pi x)\big]_0^2-\dfrac{1}{2}\left[\dfrac{1}{n\pi}\big[x\sin(n\pi x)\big]_0^2-\dfrac{1}{n\pi}\int\limits_0^2\sin(n\pi x)\ dx\right]$

$\displaystyle\small\longrightarrow a_n=0$

Finding $\displaystyle\smallb_n,$

$\displaystyle\small\longrightarrow b_n=\dfrac{1}{L}\int\limits_N^{N+2L}f(x)\sin\left(\dfrac{n\pi}{L}\,x\right)\ dx$

$\displaystyle\small\longrightarrow b_n=\int\limits_0^2\left(\dfrac{\pi-x}{2}\right)\sin(n\pi x)\ dx$

$\displaystyle\small\longrightarrow b_n=\dfrac{\pi}{2}\int\limits_0^2\sin(n\pi x)\ dx-\dfrac{1}{2}\int\limits_0^2x\sin(n\pi x)\ dx$

$\displaystyle\small\longrightarrow b_n=-\dfrac{\pi}{2n\pi}\big[\cos(n\pi x)\big]_0^2-\dfrac{1}{2}\left[-\dfrac{1}{n\pi}\big[x\cos(n\pi x)\big]_0^2+\dfrac{1}{n\pi}\int\limits_0^2\cos(n\pi x)\ dx\right]$

$\displaystyle\small\longrightarrow b_n=\dfrac{1}{n\pi}$

Hence,

$\displaystyle\small\longrightarrow f(x)=\dfrac{a_0}{2}+\sum_{n=1}^\infty\left(a_n\cos\left(\dfrac{n\pi}{L}\,x\right)+b_n\sin\left(\dfrac{n\pi}{L}\,x\right)\right)$

$\displaystyle\small\text{ \longrightarrow\underline{\underline{f(x)=\dfrac{\pi-1}{2}+\dfrac{1}{\pi}\sum_{n=1}^\infty\dfrac{\sin\left(n\pi x\right)}{n}}} }$

or,

$\displaystyle\small\text{ \longrightarrow\underline{\underline{f(x)=\dfrac{\pi-1}{2}+\dfrac{1}{\pi}\left(\sin\left(\pi x\right)+\dfrac{\sin\left(2\pi x\right)}{2}+\dfrac{\sin\left(3\pi x\right)}{3}+\dfrac{\sin\left(4\pi x\right)}{4}+\,\dots\right)}} }$

Answer 2

Step-by-step explanation:

The Fourier series of a function f(x) in the interval [N, N+2L] or (N, N+2L) is given by,

\displaystyle\small\text{$\longrightarrow f(x)=\dfrac{a_0}{2}+\sum_{n=1}^\infty\left(a_n\cos\left(\dfrac{n\pi}{L}\,x\right)+b_n\sin\left(\dfrac{n\pi}{L}\,x\right)\right)$}⟶f(x)=

2

a

0

+∑

n=1

∞

(a

n

cos(

L

nπ

x)+b

n

sin(

L

nπ

x))

where,

\displaystyle\small\text{$a_0=\dfrac{1}{L}\int\limits_N^{N+2L}f(x)\ dx$}a

0

=

L

1

N

∫

N+2L

f(x) dx

\displaystyle\small\text{$a_n=\dfrac{1}{L}\int\limits_N^{N+2L}f(x)\cos\left(\dfrac{n\pi}{L}\,x\right)\ dx$}a

n

=

L

1

N

∫

N+2L

f(x)cos(

L

nπ

x) dx

\displaystyle\small\text{$b_n=\dfrac{1}{L}\int\limits_N^{N+2L}f(x)\sin\left(\dfrac{n\pi}{L}\,x\right)\ dx$}b

n

=

L

1

N

∫

N+2L

f(x)sin(

L

nπ

x) dx

Here,

\displaystyle\small\text{$f(x)=\dfrac{\pi-x}{2}$}f(x)=

2

π−x

and the interval is (0, 2), i.e., N = 0 and L = 1.

Finding \displaystyle\small\text{$a_0,$}a

0

,

\displaystyle\small\text{$\longrightarrow a_0=\dfrac{1}{L}\int\limits_N^{N+2L}f(x)\ dx$}⟶a

0

=

L

1

N

∫

N+2L

f(x) dx

\displaystyle\small\text{$\longrightarrow a_0=\int\limits_0^2\left(\dfrac{\pi-x}{2}\right)\ dx$}⟶a

0

=

0

∫

2

(

2

π−x

) dx

\displaystyle\small\text{$\longrightarrow a_0=\dfrac{\pi}{2}\big[x\big]_0^2-\dfrac{1}{4}\left[x^2\right]_0^2$}⟶a

0

=

2

π

[x]

0

2

−

4

1

[x

2

]

0

2

\displaystyle\small\text{$\longrightarrow a_0=\pi-1$}⟶a

0

=π−1

FInding \displaystyle\small\text{$a_n,$}a

n

,

\displaystyle\small\text{$\longrightarrow a_n=\dfrac{1}{L}\int\limits_N^{N+2L}f(x)\cos\left(\dfrac{n\pi}{L}\,x\right)\ dx$}⟶a

n

=

L

1

N

∫

N+2L

f(x)cos(

L

nπ

x) dx

\displaystyle\small\text{$\longrightarrow a_n=\int\limits_0^2\left(\dfrac{\pi-x}{2}\right)\cos(n\pi x)\ dx$}⟶a

n

=

0

∫

2

(

2

π−x

)cos(nπx) dx

\displaystyle\small\text{$\longrightarrow a_n=\dfrac{\pi}{2}\int\limits_0^2\cos(n\pi x)\ dx-\dfrac{1}{2}\int\limits_0^2x\cos(n\pi x)\ dx$}⟶a

n

=

2

π

0

∫

2

cos(nπx) dx−

2

1

0

∫

2

xcos(nπx) dx

\displaystyle\small\text{$\longrightarrow a_n=\dfrac{\pi}{2n\pi}\big[\sin(n\pi x)\big]_0^2-\dfrac{1}{2}\left[\dfrac{1}{n\pi}\big[x\sin(n\pi x)\big]_0^2-\dfrac{1}{n\pi}\int\limits_0^2\sin(n\pi x)\ dx\right]$}⟶a

n

=

2nπ

π

[sin(nπx)]

0

2

−

2

1

[

nπ

1

[xsin(nπx)]

0

2

−

nπ

1

0

∫

2

sin(nπx) dx]

\displaystyle\small\text{$\longrightarrow a_n=0$}⟶a

n

=0

Finding \displaystyle\small\text{$b_n,$}b

n

,

\displaystyle\small\text{$\longrightarrow b_n=\dfrac{1}{L}\int\limits_N^{N+2L}f(x)\sin\left(\dfrac{n\pi}{L}\,x\right)\ dx$}⟶b

n

=

L

1

N

∫

N+2L

f(x)sin(

L

nπ

x) dx

\displaystyle\small\text{$\longrightarrow b_n=\int\limits_0^2\left(\dfrac{\pi-x}{2}\right)\sin(n\pi x)\ dx$}⟶b

n

=

0

∫

2

(

2

π−x

)sin(nπx) dx

\displaystyle\small\text{$\longrightarrow b_n=\dfrac{\pi}{2}\int\limits_0^2\sin(n\pi x)\ dx-\dfrac{1}{2}\int\limits_0^2x\sin(n\pi x)\ dx$}⟶b

n

=

2

π

0

∫

2

sin(nπx) dx−

2

1

0

∫

2

xsin(nπx) dx

\displaystyle\small\text{$\longrightarrow b_n=-\dfrac{\pi}{2n\pi}\big[\cos(n\pi x)\big]_0^2-\dfrac{1}{2}\left[-\dfrac{1}{n\pi}\big[x\cos(n\pi x)\big]_0^2+\dfrac{1}{n\pi}\int\limits_0^2\cos(n\pi x)\ dx\right]$}⟶b

n

=−

2nπ

π

[cos(nπx)]

0

2

−

2

1

[−

nπ

1

[xcos(nπx)]

0

2

+

nπ

1

0

∫

2

cos(nπx) dx]

\displaystyle\small\text{$\longrightarrow b_n=\dfrac{1}{n\pi}$}⟶b

n

=

nπ

1

Hence,

\displaystyle\small\text{$\longrightarrow f(x)=\dfrac{a_0}{2}+\sum_{n=1}^\infty\left(a_n\cos\left(\dfrac{n\pi}{L}\,x\right)+b_n\sin\left(\dfrac{n\pi}{L}\,x\right)\right)$}⟶f(x)=

2

a

0

+∑

n=1

∞

(a

n

cos(

L

nπ

x)+b

n

sin(

L

nπ

x))

\displaystyle\small\text{$\longrightarrow\underline{\underline{f(x)=\dfrac{\pi-1}{2}+\dfrac{1}{\pi}\sum_{n=1}^\infty\dfrac{\sin\left(n\pi x\right)}{n}}}$}⟶

f(x)=

2

π−1

+

π

1

∑

n=1

∞

n

sin(nπx)

or,

\displaystyle\small\text{$\longrightarrow\underline{\underline{f(x)=\dfrac{\pi-1}{2}+\dfrac{1}{\pi}\left(\sin\left(\pi x\right)+\dfrac{\sin\left(2\pi x\right)}{2}+\dfrac{\sin\left(3\pi x\right)}{3}+\dfrac{\sin\left(4\pi x\right)}{4}+\,\dots\right)}}$}⟶

f(x)=

2

π−1

+

π

1

(sin(πx)+

2

sin(2πx)

+

3

sin(3πx)

+

4

sin(4πx)

+…)